Question Number 163902 by HongKing last updated on 11/Jan/22
$$\sqrt{\mathrm{x}!^{\boldsymbol{\mathrm{x}}!} }\:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}!} \:\:=\:\mathrm{x}!^{\mathrm{3}} \:\:+\:\:\mathrm{10x}!\:\:+\:\:\mathrm{4} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Answered by MJS_new last updated on 11/Jan/22
$${x}!\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{try}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}... \\ $$$$\Rightarrow\:{x}=\mathrm{3} \\ $$